# Bessel differential equation, Bessel functions and variable sign change

The homogeneous Bessel differential equation

$$x^2 f''(x) + xf'(x) + (x^2 - \nu^2) f(x) = 0$$

does not change if $$x$$ is substituted with $$-x$$. So, it could be expected that it is the same as regards the solutions: $$J_{\nu} (-x) = J_{\nu}(x)$$. However, it is apparently not the case!

The Bessel function of the first kind $$J_{\nu} (x)$$ is even or odd respectively when $$\nu$$ is integer and even or odd. It is the same for: the modified Bessel function of the first kind $$I_{\nu} (x)$$ (which is proportional to $$J_{\nu} (x)$$), the Bessel function of the second kind $$Y_{\nu} (x)$$ (which depends on $$J_{\nu} (x)$$ and $$J_{-\nu} (x)$$) and the modified Bessel function of the second kind $$K_{\nu} (x)$$ (which depends on $$I_{\nu} (x)$$ and $$I_{-\nu} (x)$$).

When $$\nu$$ is complex, instead:

$$J_{\nu}(-x) = e^{j \pi \nu} J_{\nu}(x)$$

as well as

$$I_{\nu}(-x) = e^{j \pi \nu} I_{\nu}(x)$$

$$e^{j \pi \nu}$$ are in general complex values.

How is it possible that $$J_{\nu}(-x) \neq J_{\nu}(x)$$ as well as for the other three functions? If the equation does not change when $$x \rightarrow -x$$, how can instead the solutions change?

From the symmetry you only know that when $$f(x)$$ is a solution, then also $$f(-x)$$ is a solution. Then also the linear combinations $$f(x)+f(-x)$$ and $$f(x)-f(-x)$$ are solutions because of the linearity of the ODE, which are, if not zero, even and odd functions respectively. As odd and even are obviously linearly independent, one can try to get an even-odd pair of basis solutions.
• "From the symmetry you only know that when $f(x)$ is a solution, then also $f(−x)$ is a solution". Ok! But I can not figure out: why? – BowPark Nov 9 at 9:41
• To cite your question: "does not change if $x$ is substituted with $−x$", just set $g(x)=f(-x)$ to get $g'(x)=-f'(-x)$, $g''(x)=f''(-x)$ and if you insert this into the equation for $g$, you get $$(-x)^2f''(-x)+(-x)f'(-x)+((-x)^2-ν^2)f(-x)=0$$ which is true if $f$ is a solution. – LutzL Nov 9 at 9:49